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Veterans Upward Bound
Algebra I Concepts - Honors
Brenda Meery
Kaitlyn Spong
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C HAPTER
4
Exponents
Chapter Outline
4.1
Z ERO AND N EGATIVE E XPONENTS
4.2
P RODUCT R ULES FOR E XPONENTS
4.3
Q UOTIENT R ULES FOR E XPONENTS
4.4
P OWER R ULE FOR E XPONENTS
4.5
E XPONENTIAL E XPRESSIONS
4.6
S CIENTIFIC N OTATION
Introduction
Here you’ll learn all about exponents in algebra. You will learn the properties of exponents and how to simplify
exponential expressions. You will learn how exponents can help you write very large or very small numbers with
scientific notation. You will also learn how to solve different types of exponential equations where the variable
appears as the exponent or the base. Finally, you will explore different types of exponential functions of the form
x
x
y = bx , y = abx , y = ab c , and y = ab c + d as well as applications of exponential functions.
314
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Chapter 4. Exponents
4.1 Zero and Negative Exponents
Here you’ll learn how to work with zero and negative exponents.
How can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent?
Watch This
Khan Academy Negative Exponent Intuition
MEDIA
Click image to the left for more content.
Guidance
Zero Exponent
Recall that
am
= am−n
an
. If m = n, then the following would be true:
am
= am−n = a0
an
33
= 33−3 = 30
33
However, any quantity divided by itself is equal to one. Therefore,
general:
33
33
= 1 which means 30 = 1. This is true in
a0 = 1 if a 6= 0.
Note that if a = 0, 00 is not defined.
Negative Exponents
42 × 4−2 = 42+(−2) = 40 = 1
Therefore:
315
4.1. Zero and Negative Exponents
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42 × 4−2 = 1
42 × 4−2
1
= 2
2
4
4
2
−2
4 ×4
1
= 2
2
4
4
4−2 =
Divide both sides by 42 .
Simplify the equation.
1
42
This is true in general and creates the following laws for negative exponents:
•
a−m =
1
am
•
1
a−m
= am
These laws for negative exponents can be expressed in many ways:
• If a term has a negative exponent, write it as 1 over the term with a positive exponent. For example: a−m = a1m
1
and a−m
= am
−2
2
• If a term has a negative exponent, write the reciprocal with a positive exponent. For example: 23
= 32
−m
and a−m = a 1 = a1m
• If the term is a factor in the numerator with a negative exponent, write it in the denominator with a positive
n
exponent. For example: 3x−3 y = 3y
and a−m bn = a1m (bn ) = abm
x3
• If the term is a factor in the denominator with a negative exponent, write it in the numerator with a positive
3
n
m
exponent. For example: x2x−2 = 2x3 (x2 ) and ab−m = bn a1 = bn am
These ways for understanding negative exponents provide shortcuts for arriving at solutions without doing tedious
calculations. The results will be the same.
Example A
Evaluate the following using the laws of exponents.
3 −2
4
Solution:
There are two methods that can be used to evaluate the expression.
Method 1: Apply the negative exponent rule
a−m =
316
1
am
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Chapter 4. Exponents
−2
3
=
4
1
3 2
Write the expression with a positive exponent by applying a−m =
4
1
1
= 32
2
3
42
4
1
=
32
42
1
1
a n
b
=
an
bn
=
an
bn
Evaluate the powers.
9
16
= 1÷
9
16
Apply the law of exponents for raising a quotient to a power.
1
.
am
9
16
Divide
9
16 16
= 1×
=
16
9
9
−2
3
16
=
4
9
1÷
Method 2: Apply the shortcut and write the reciprocal with a positive exponent.
−2 2
3
4
=
4
3
2
4
42
= 2
3
3
42 16
=
32
9
−2
3
16
=
4
9
Write the reciprocal with a positive exponent.
Apply the law of exponents for raising a quotient to a power.
a n
b
Simplify.
Applying the shortcut facilitates the process for obtaining the solution.
Example B
State the following using only positive exponents: (If possible, use shortcuts)
i) y−6
−3
ii) ab
iii)
x5
y−4
iv) a2 × a−5
Solutions:
i)
y−6
y−6 =
Write the expression with a positive exponent by applying
a−m =
1
.
am
1
y6
317
4.1. Zero and Negative Exponents
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ii)
a −3
Write the reciprocal with a positive exponent.
b
a −3 b 3
=
b
a
3
b
b3
= 3
a
a
a −3 b3
= 3
b
a
a n
Apply the law of exponents for raising a quotient to a power.
b
=
an
bn
iii)
x5
y−4
Apply the negative exponent rule.
x5
= x5
y−4
y4
1
1
a−m
= am
Simplify.
x5
= x 5 y4
y−4
iv)
a2 × a−5
Apply the product rule for exponents am × an = am+n .
a2 × a−5 = a2+(−5)
Simplify.
a2+(−5) = a−3
Write the expression with a positive exponent by applying a−m =
a−3 =
1
a3
a2 × a−5 =
1
a3
Example C
Evaluate the following:
7−2 +7−1
7−3 +7−4
Solution:
There are two methods that can be used to evaluate the problem.
Method 1: Work with the terms in the problem in exponential form.
Numerator:
318
1
.
am
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Chapter 4. Exponents
7−2 =
1
1
and 7−1 =
2
7
7
Apply the definition a−m =
1 1
+
72 7 1 1 7
+
72 7 7
7
1+7
8
1
+ 2= 2 = 2
2
7
7
7
7
1
am
A common denominator is needed to add the fractions.
Multiply
1
7
by to obtain the common denominator of 72
7
7
Add the fractions.
Denominator:
1
1
and 7−4 = 4
3
7
7
1
1
+
73 74
7 1
1
+ 4
3
7 7
7
1
1+7
8
7
+ 4= 4 = 4
4
7
7
7
7
7−3 =
Apply the definition a−m =
1
am
A common denominator is needed to add the fractions.
Multiply
1
7
by to obtain the common denominator of 74
3
7
7
Add the fractions.
Numerator and Denominator:
8
8
÷ 4
2
7
7
8 74
×
72
8
74 74
8
× = 2 = 72 = 49
72
7
8
Divide the numerator by the denominator.
Multiply by the reciprocal.
Simplify.
7−2 + 7−1
= 49
7−3 + 7−4
Method 2: Multiply the numerator and the denominator by 74 . This will change all negative exponents to
positive exponents. Apply the product rule for exponents and work with the terms in exponential form.
7−2 + 7−1
7−3 + 7−4
4 −2
7 7 + 7−1
74 7−3 + 7−4
72 + 73
71 + 70
49 + 343 392
=
= 49
7+1
8
Apply the distributive property with the product rule for exponents.
Evaluate the numerator and the denominator.
7−2 + 7−1
= 49
7−3 + 7−4
Whichever method is used, the result is the same.
319
4.1. Zero and Negative Exponents
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Concept Problem Revisited
m
By the quotient rule for exponents, xxm = xm−m = x0 . Since anything divided by itself is equal to 1 (besides 0),
Therefore, x0 = 1 as long as x 6= 0.
Also by the quotient rule for exponents,
you would have
x2
x5
=
x·x
x·x·x·x·x
=
1
.
x3
x2
x5
xm
xm
= 1.
= x2−5 = x−3 . If you were to expand and reduce the original expression
Therefore, x−3 =
1
.
x3
This generalizes to x−a =
1
xa .
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2 × 2 × 2.
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
Guided Practice
1. Use the laws of exponents to simplify the following: (−3x2 )3 (9x4 y)−2
2. Rewrite the following using only positive exponents. (x2 y−1 )2
3. Use the laws of exponents to evaluate the following: [5−4 × (25)3 ]2
Answers:
1.
320
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Chapter 4. Exponents
(−3x2 )3 (9x4 y)−2
Apply the laws of exponents (am )n = amn and a−m =
1
am
1
(−3x ) (9x y) = (−3 x )
Simplify and apply (ab)n = an bn
(9x4 y)2
1
1
6
(−33 x6 )
=
−27x
Simplify.
(9x4 y)2
(92 x8 y2 )
am
−27x6
1
6
Simplify
and
apply
the
quotient
rule
for
exponents
= am−n .
=
− 27x
(92 x8 y2 )
81x8 y2
an
2 3
4
−2
3 6
1x−2
−27x6
=
−
81x8 y2
3y2
Apply the negative exponent rule a−m =
(−3x2 )3 (9x4 y)−2 = −
1
am
1
3x2 y2
2.
(x2 y−1 )2 = x4 y−2
=
x4
y2
3.
[5−4 × (25)3 ]2
Try to do this one by applying the laws of exponents.
[5
−4
3 2
× (25) ] = [5
−4
× (5 ) ]
[5
−4
2 3 2
−4
× 56 ]2
× (5 ) ] = [5
2 3 2
[5−4 × 56 ]2 = (52 )2
(52 )2 = 54
54 = 625
[5−4 × (25)3 ]2 = 54 = 625
Practice
Evaluate each of the following expressions:
2 0
3
−2
− 25
(−3)−3
1. −
2.
3.
−2
4. 6 × 12
5. 7−4 × 74
Rewrite the following using positive exponents only. Simplify where possible.
6. (4wx−2 y3 z−4 )3
321
4.1. Zero and Negative Exponents
2 3 −2
7. da−2bbcc −6
8. x−2 (x2 − 1)
9. m4 (m2 + m − 5m−2 )
−2 −2
10. xx−1 yy−1
−2 3 −4 −7
y
11. xy4
x6
12.
13.
14.
15.
322
(x−2 y4 )2
(x5 y−3 )4
(3xy2 )3
(3x2 y)4
0
x2 y−25 z5
−12.4x3 y
−2 5 −2 −3
y
x
y3
x4
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Chapter 4. Exponents
4.2 Product Rules for Exponents
Here you’ll learn how to multiply two terms with the same base and how to find the power of a product.
Suppose you have the expression:
x·x·x·x·x·x·x·x·x·y·y·y·y·y·x·x·x·x
How could you write this expression in a more concise way?
Watch This
MEDIA
Click image to the left for more content.
James Sousa: Exponential Notation
Guidance
In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as
powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself.
For example:
• x3 = x · x · x
• 24 = 2 · 2 · 2 · 2 = 16.
There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know
so that you can work with expressions and equations that involve exponents more easily. Here you will learn two
rules that have to do with exponents and products.
RULE: To multiply two terms with the same base, add the exponents.
am × an = (a × a × . . . × a) (a × a × . . . × a)
←−−−−−−−−−→ ←−−−−−−−−−→
↓
↓
m factors
m
n
m
n
n factors
a × a = (a × a × a . . . × a)
←−−−−−−−−−−→
↓
m + n factors
a ×a = a
m+n
323
4.2. Product Rules for Exponents
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RULE: To raise a product to a power, raise each of the factors to the power.
(ab)n = (ab) × (ab) × . . . × (ab)
←−−−−−−−−−−−−−−→
↓
n factors
n
(ab) = (a × a × . . . × a) × (b × b × . . . × b)
←−−−−−−−−−→ ←−−−−−−−−−→
↓
↓
n factors
n
n factors
n n
(ab) = a b
Example A
Simplify 32 × 33 .
Solution:
32 × 33
The base is 3.
32+3
Keep the base of 3 and add the exponents.
3
5
This answer is in exponential form.
The answer can be taken one step further. The base is numerical so the term can be evaluated.
35 = 3 × 3 × 3 × 3 × 3
35 = 243
32 × 33 = 35 = 243
Example B
Simplify (x3 )(x6 ).
Solution:
(x3 )(x6 )
x
3+6
x9
(x3 )(x6 ) = x9
Example C
Simplify y5 · y2 .
324
The base is x.
Keep the base of x and add the exponents.
The answer is in exponential form.
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Chapter 4. Exponents
Solution:
y5 · y2
5+2
The base is y.
y
Keep the base of y and add the exponents.
y7
The answer is in exponential form.
y5 · y2 = y7
Example D
Simplify 5x2 y3 · 3xy2 .
Solution:
5x2 y3 · 3xy2
The bases are x and y.
15(x2 y3 )(xy2 )
Multiply the coefficients - 5 × 3 = 15. Keep the base of x and y and add
the exponents of the same base. If a base does not have a written
exponent, it is understood as 1.
15x
2+1 3+2
y
15x3 y5
The answer is in exponential form.
5x2 y3 · 3xy2 = 15x3 y5
Concept Problem Revisited
x · x · x · x · x · x · x · x · x · y · y · y · y · y · x · x · x · x can be rewritten as x9 y5 x4 . Then, you can use the rules of exponents to
simplify the expression to x13 y5 . This is certainly much quicker to write!
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2 × 2 × 2.
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
325
4.2. Product Rules for Exponents
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Guided Practice
Simplify each of the following expressions.
1. (−3x)2
2. (5xy)3
3. (23 · 32 )2
Answers:
1. 9x2 . Here are the steps:
(−3x)2 = (−3)2 · (x)2
= 9x2
2. 125x3 y3 . Here are the steps:
(5x2 y4 )3 = (5)3 · (x)3 · (y)3
= 125x3 y3
3. 5184. Here are the steps:
(23 · 32 )2 = (8 · 9)2
= (72)2
= 5184
OR
(23 · 32 )2 = (8 · 9)2
= 82 · 92
= 64 · 81
= 5184
Practice
Simplify each of the following expressions, if possible.
1.
2.
3.
4.
5.
6.
7.
8.
326
42 × 44
x4 · x12
(3x2 y4 )(9xy5 z)
(2xy)2 (4x2 y3 )
(3x)5 (2x)2 (3x4 )
x3 y2 z · 4xy2 z7
x2 y3 + xy2
(0.1xy)4
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9.
10.
11.
12.
Chapter 4. Exponents
(xyz)6
2x4 (x2 − y2 )
3x5 − x2
3x8 (x2 − y4 )
Expand and then simplify each of the following expressions.
13. (x5 )3
14. (x6 )8
15. (xa )b Hint: Look for a pattern in the previous two problems.
327
4.3. Quotient Rules for Exponents
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4.3 Quotient Rules for Exponents
Here you’ll learn how to divide two terms with the same base and find the power of a quotient.
Suppose you have the expression:
x·x·x·x·x·x·x·x·x·y·y·y·y·y
x·x·x·x·x·x·y·y·y
How could you write this expression in a more concise way?
Watch This
MEDIA
Click image to the left for more content.
James Sousa: Simplify Exponential Expressions- Quotient Rule
Guidance
In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as
powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself.
For example:
• x3 = x · x · x
• 24 = 2 · 2 · 2 · 2 = 16.
There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know
so that you can work with expressions and equations that involve exponents more easily. Here you will learn two
rules that have to do with exponents and quotients.
RULE: To divide two powers with the same base, subtract the exponents.
328
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Chapter 4. Exponents
m factors
↑
←
−
−
−
−
−
−−−−→
am (a × a × . . . × a)
=
m > n; a 6= 0
an
(a × a × . . . × a)
←−−−−−−−−−→
↓
n factors
am
an
= (a × a × . . . × a)
←−−−−−−−−−→
↓
m − n factors
am
an
= am−n
RULE: To raise a quotient to a power, raise both the numerator and the denominator to the power.
a n
b
=
a a
a
× ×...×
b b
b
←−−−−−−−−−→
↓
n factors
n factors
↑
←
−
−
−
−
−
−
−→
a n (a × a × . .−.−
× a)
=
b
(b × b × . . . × b)
←−−−−−−−−−→
↓
a n
b
n factors
an
= n (b 6= 0)
b
Example A
Simplify 27 ÷ 23 .
Solution:
27 ÷ 23
The base is 2.
2
7−3
Keep the base of 2 and subtract the exponents.
2
4
The answer is in exponential form.
The answer can be taken one step further. The base is numerical so the term can be evaluated.
329
4.3. Quotient Rules for Exponents
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24 = 2 × 2 × 2 × 2
24 = 16
27 ÷ 23 = 24 = 16
Example B
Simplify
x8
.
x2
Solution:
x8
x2
x8−2
x
The base is x.
Keep the base of x and subtract the exponents.
6
The answer is in exponential form.
x8
= x6
x2
Example C
Simplify
16x5 y5
.
4x2 y3
Solution:
16x5 y5
4x2 y3
5 5
x y
4 2 3
x y
The bases are x and y.
Divide the coefficients - 16 ÷ 4 = 4. Keep the base of x and y and
subtract the exponents of the same base.
4x5−2 y5−3
4x3 y2
Concept Problem Revisited
x·x·x·x·x·x·x·x·x·y·y·y·y·y
x·x·x·x·x·x·y·y·y
can be rewritten as
x9 y5
x6 y3
and then simplified to x3 y2 .
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
330
www.ck12.org
Chapter 4. Exponents
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2 × 2 × 2.
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
Guided Practice
Simplify each of the following expressions.
2
1. 32
3
2. 6x
2
3. 3x
4y
Answers:
2
2
1. 23 = 232 = 94
3
3
x3
2. 6x = 6x3 = 216
2
2 2
9x2
3. 3x
= 342 xy2 = 16y
2
4y
Practice
Simplify each of the following expressions, if possible.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
2 6
5
4 3
7
4
x
y
20x4 y5
5x2 y4
42x2 y8 z2
4
6xyz3
3x
4y
72x2 y4
8x2 y3
x 5
4
24x14 y8
3x5 y7
72x3 y9
6
24xy3
7
y
20x12
−5x8
13. Simplify using the laws of exponents:
23
25
14. Evaluate the numerator and denominator separately and then simplify the fraction:
15. Use your result from the previous problem to determine the value of a:
23
25
=
23
25
1
2a
331
4.3. Quotient Rules for Exponents
16. Use your results from the previous three problems to help you evaluate 2−4 .
332
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Chapter 4. Exponents
4.4 Power Rule for Exponents
Here you’ll learn how to find the power of a power.
Can you simplify an expression where an exponent has an exponent? For example, how would you simplify [(23 )2 ]4 ?
Watch This
MEDIA
Click image to the left for more content.
James Sousa: Properties of Exponents
Guidance
In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as
powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself.
For example:
• x3 = x · x · x
• 24 = 2 · 2 · 2 · 2 = 16.
There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know
so that you can work with expressions and equations that involve exponents more easily. Here you will learn a rule
that has to do with raising a power to another power.
RULE: To raise a power to a new power, multiply the exponents.
(am )n = (a × a × . . . × a)n
←−−−−−−−−−−→
↓
m factors
m n
(a ) = (a × a × . . . × a) × (a × a × . . . × a) (a × a × . . . × a)
←−−−−−−−−−→ ←−−−−−−−−−→ ←−−−−−−−−−→
↓
↓
↓
m factors
m factors
m factors
←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
n times
(am )n = a × a × a . . . × a
←−−−−−−−−−→
mn factors
(am )n = amn
333
4.4. Power Rule for Exponents
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Example A
Evaluate (23 )2 .
Solution: (23 )2 = 26 = 64.
Example B
Simplify (x7 )4 .
Solution: (x7 )4 = x28 .
Example C
Evaluate (32 )3 .
Solution: (32 )3 = 36 = 729.
Example D
Simplify (x2 y4 )2 · (xy4 )3 .
Solution: (x2 y4 )2 · (xy4 )3 = x4 y8 · x3 y12 = x7 y20 .
Concept Problem Revisited
[(23 )2 ]4 = [26 ]4 = 224 . Notice that the power rule applies even when a number has been raised to more than one
power. The overall exponent is 24 which is 3 · 2 · 4.
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2 × 2 × 2.
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
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Chapter 4. Exponents
Guided Practice
You know you can rewrite 24 as 2 × 2 × 2 × 2 and then calculate in order to find that
24 = 16
. This concept can also be reversed. To write 32 as a power of 2, 32 = 2 × 2 × 2 × 2 × 2. There are 5 twos; therefore,
32 = 25
. Use this idea to complete the following problems.
1. Write 81 as a power of 3.
2. Write (9)3 as a power of 3.
3. Write (43 )2 as a power of 2.
Answers:
1. 81 = 3 × 3 = 9 × 3 = 27 × 3 = 81
There are 4 threes. Therefore
81 = 34
2. 9 = 3 × 3 = 9
There are 2 threes. Therefore
9 = 32
.
(32 )3
Apply the law of exponents for power to a power-multiply the exponents.
32×3 = 36
Therefore
(9)3 = 36
3. 4 = 2 × 2 = 4
There are 2 twos. Therefore
4 = 22
(22 )3
2
Apply the law of exponents for power to a power-multiply the exponents.
22×3 = 26
(26 )2 Apply the law of exponents for power to a power-multiply the exponents.
26×2 = 212
Therefore
(43 )2 = 212
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4.4. Power Rule for Exponents
Practice
Simplify each of the following expressions.
1.
2.
3.
4 5
x
y3
(5x2 y4 )5
(5xy2 )3
x 8 y9
(x2 y)3
(x2 y4 )3
4.
5. (3x2 )2 · (4xy4 )2
6. (2x3 y5 )(5x2 y)3
7. (x4 y6 z2 )2 (3xyz)3
2 4
x
8. 2y
3
9.
10.
11.
12.
13.
14.
15.
16.
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(4xy3 )4
(2xy2 )3
True or false: (x2 + y3 )2 = x4 + y6
True or false: (x2 y3 )2 = x4 y6
Write 64 as a power of 4.
Write (16)3 as a power of 2.
Write (94 )2 as a power of 3.
Write (81)2 as a power of 3.
Write (253 )4 as a power of 5.
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Chapter 4. Exponents
4.5 Exponential Expressions
Here you’ll learn how to use all of the laws of exponents to simplify and evaluate exponential expressions.
Can you simplify the following expression so that it has only positive exponents?
8x3 y−2
(−4a2 b4 )−2
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Guidance
The following table summarizes all of the rules for exponents.
Laws of Exponents
If a ∈ R, a ≥ 0 and m, n ∈ Q, then
1.
2.
3.
4.
5.
6.
7.
8.
am × an = am+n
am
m−n (if m > n, a 6= 0)
an = a
m
n
(a ) = amn
n
n n
(ab)
= aan b
a n
= bn (b 6= 0)
b
a0 = 1 (a 6= 0)
a−m = a1m
√
√ m
m
n
a n = am = n a
Example A
1
Evaluate 81− 4 .
Solution: First, rewrite with a positive exponent:
1
1
1 4
81− 4 = 11 = 81
.
81 4
Next, evaluate the fractional exponent:
r
1
1
4
1 4
=
=1
81
81 3
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4.5. Exponential Expressions
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Example B
Simplify (4x3 y)(3x5 y2 )4 .
Solution:
(4x3 y)(3x5 y2 )4 = (4x3 y)(81x20 y8 )
= 324x23 y9
Example C
Simplify
x−2 y
x 4 y3
−2
.
Solution:
x−2 y
x 4 y3
−2
=
x 4 y3
x−2 y
2
= (x6 y2 )2
= x12 y4
Concept Problem Revisited
8x3 y−2
= (8x3 y−2 )(−4x2 y4 )2
(−4x2 y4 )−2
= (8x3 y−2 )(16x4 y8 )
= 8 · 16 · x3 · x4 · y−2 · y8
= 128x7 y6
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2 × 2 × 2.
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
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Chapter 4. Exponents
Guided Practice
Use the laws of exponents to simplify each of the following:
1. (−2x)5 (2x2 )
2. (16x10 ) 43 x5
3.
(x15 )(x24 )(x25 )
(x7 )8
Answers:
1. (−2x)5 (2x2 ) = (−32x5 )(2x2 ) = −64x7
2. (16x10 ) 43 x5 = 12x15
3.
(x15 )(x24 )(x25 )
(x7 )8
=
x64
x56
= x8
Practice
Simplify each expression.
1. (x10 )(x10 )
2. (7x3 )(3x7 )
3. (x3 y2 )(xy3 )(x5 y)
3
2)
4. (x(x)(x
4)
5.
x2
x−3
x 6 y8
x4 y−2
(2x12 )3
6.
7.
8. (x5 y10 )7
10 3
9. 2x
3y20
Express each of the following as a power of 3. Do not evaluate.
10.
11.
12.
13.
14.
(33 )5
(39 )(33 )
(9)(37 )
94
(9)(272 )
Apply the laws of exponents to evaluate each of the following without using a calculator.
15.
16.
17.
18.
19.
(23 )(22 )
66 ÷ 65
−(32 )3
(12 )3 + (13 )2
1 6
1 8
÷
3
3
Use the laws of exponents to simplify each of the following.
20. (4x)2
21. (−3x)3
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4.5. Exponential Expressions
22.
23.
24.
25.
26.
(x3 )4
(3x)(x7 )
(5x)(4x4 )
(−3x2 )(−6x3 )
(10x8 ) ÷ (2x4 )
Simplify each of the following using the laws of exponents.
1
1
27. 5 2 × 5 3
4 8 12 14
28. (d
se f )
1 √
xy
4 y2
29.
2
x3
1
30. (32a20 b−15 ) 5
2
31. (729x12 y−6 ) 3
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Chapter 4. Exponents
4.6 Scientific Notation
Here you’ll learn about scientific notation.
Very large and very small quantities and measures are often used to provide information in magazines, textbooks,
television, newspapers and on the Internet. Some examples are:
• The distance between the sun and Neptune is 4,500,000,000 km.
• The diameter of an electron is approximately 0.00000000000022 inches.
Scientific notation is a convenient way to represent such numbers. How could you write the numbers above using
scientific notation?
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Guidance
To represent a number in scientific notation means to express the number as a product of two factors: a number
between 1 and 10 (including 1) and a power of 10. A positive real number ’x’ is said to be written in scientific
notation if it is expressed as
x = a × 10n
where
1 ≤ a < 10 and n ∈ Z.
In other words, a number in scientific notation is a single nonzero digit followed by a decimal point and other digits,
all multiplied by a power of 10.
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4.6. Scientific Notation
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When working with numbers written in scientific notation, you can use the following rules. These rules are proved
by example in Example B and Example C.
(A × 10n ) + (B × 10n ) = (A + B) × 10n
(A × 10n ) − (B × 10n ) = (A − B) × 10n
(A × 10m ) × (B × 10n ) = (A × B) × (10m+n )
(A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n )
Example A
Write the following numbers using scientific notation:
i) 2,679,000
ii) 0.00005728
Solutions:
i)
2, 679, 000 = 2.679 × 1, 000, 000
2.679 × 1, 000, 000 = 2.679 × 106
The exponent, n = 6, represents the decimal point that is 6 places to the right of the standard position of the
decimal point.
ii)
0.00005728 = 5.728 × 0.00001
1
5.728 × 0.00001 = 5.728 ×
100, 000
1
1
5.728 ×
= 5.728 × 5
100, 000
10
1
5.728 ×
= 5.728 × 10−5
100, 000
The exponent, n = −5, represents the decimal point that is 5 places to the left of the standard position of the
decimal point.
One advantage of scientific notation is that calculations with large or small numbers can be done by applying the
laws of exponents.
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Chapter 4. Exponents
Example B
Complete the following table.
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4.6. Scientific Notation
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TABLE 4.1:
Expression in Scientific
Notation
1.3 × 105 + 2.5 × 105
3.7 × 10−2 + 5.1 × 10−2
4.6 × 104 − 2.2 × 104
7.9 × 10−2 − 5.4 × 10−2
Expression in Standard
Form
Result
Form
in
Standard
Result in Scientific Notation
Result in
Form
380,000
0.088
24,000
0.025
Standard
Result in Scientific Notation
3.8 × 105
8.8 × 10−2
2.4 × 104
2.5 × 10−2
Solution:
TABLE 4.2:
Expression in Scientific
Notation
1.3 × 105 + 2.5 × 105
3.7 × 10−2 + 5.1 × 10−2
4.6 × 104 − 2.2 × 104
7.9 × 10−2 − 5.4 × 10−2
Expression in Standard
Form
130, 000 + 250, 000
0.037 + 0.051
46, 000 − 22, 000
0.079 − 0.054
Note that the numbers in the last column have the same power of 10 as those in the first column.
Example C
Complete the following table.
TABLE 4.3:
Expression in Scientific
Notation
(3.6 × 102 ) × (1.4 × 103 )
(2.5 × 103 ) × (1.1 × 10−6 )
(4.4 × 104 ) ÷ (2.2 × 102 )
(6.8 × 10−4 ) ÷ (3.2 ×
10−2 )
Expression in Standard
Form
Result
Form
in
Standard
Result in Scientific Notation
Result in
Form
504,000
0.00275
200
0.02125
Standard
Result in Scientific Notation
5.04 × 105
2.75 × 10−3
2.0 × 102
2.125 × 10−2
Solution:
TABLE 4.4:
Expression in Scientific
Notation
(3.6 × 102 ) × (1.4 × 103 )
(2.5 × 103 ) × (1.1 × 10−6 )
(4.4 × 104 ) ÷ (2.2 × 102 )
(6.8 × 10−4 ) ÷ (3.2 ×
10−2 )
Expression in Standard
Form
360 × 1400
2500 × 0.0000011
44, 000 ÷ 220
0.00068 ÷ 0.032
Note that for multiplication, the power of 10 is the result of adding the exponents of the powers in the first column.
For division, the power of 10 is the result of subtracting the exponents of the powers in the first column.
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Chapter 4. Exponents
Example D
Calculate each of the following:
i) 4.6 × 104 + 5.3 × 105
ii) 4.7 × 10−3 − 2.4 × 10−4
iii) (7.3 × 105 ) × (6.8 × 104 )
iv) (4.8 × 109 ) ÷ (5.79 × 107 )
Solution:
i) Before the rule
(A × 10n ) + (B × 10n ) = (A + B) × 10n
can be used, one of the numbers must be rewritten so that the powers of 10 are the same.
Rewrite 4.6 × 104
4.6 × 104 = (0.46 × 101 ) × 104 The power 101 indicates the number of places to the right that the decimal point must
be moved to return 0.46 to the original number of 4.6.
(0.46 × 101 ) × 104 = 0.46 × 105 Add the exponents of the power.
Rewrite the question and substitute 4.6 × 104 with 0.46 × 105 .
0.46 × 105 + 5.3 × 105
Apply the rule
(A × 10n ) + (B × 10n ) = (A + B) × 10n
.
(0.46 × 105 ) + (5.3 × 105 ) = (0.46 + 5.3) × 105
(0.46 + 5.3) × 105 = 5.76 × 105
4.6 × 104 + 5.3 × 105 = 5.76 × 105
ii) Before the rule
(A × 10n ) − (B × 10n ) = (A − B) × 10n
can be used, one of the numbers must be rewritten so that the powers of 10 are the same.
Rewrite 4.7 × 10−3
4.7 × 10−3 = (47 × 10−1 ) × 10−3 The power 10−1 indicates the number of places to the left that the decimal point
must be moved to return 47 to the original number of 4.7.
(47 × 10−1 ) × 10−3 = 47 × 10−4 Add the exponents of the power.
Rewrite the question and substitute 4.7 × 10−3 with 47 × 10−4 .
47 × 10−4 − 2.4 × 10−4
Apply the rule
(A × 10n ) − (B × 10n ) = (A − B) × 10n
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4.6. Scientific Notation
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.
(47 × 10−4 ) − (2.4 × 10−4 ) = (47 − 2.4) × 10−4
(47 × 10−4 ) − (2.4 × 10−4 ) = 44.6 × 10−4
The answer must be written in scientific notation.
44.6 × 10−4 = (4.46 × 101 ) × 10−4
−4
4.46 × 10 × 10
Apply the law of exponents − add the exponents of the power.
−3
= 4.46 × 10
4.7 × 10−3 − 2.4 × 10−4 = 4.46 × 10−3
iii) (7.3 × 105 ) × (6.8 × 104 )
7.3 × 105 × 6.8 × 104
Apply the rule (A × 10m ) × (B × 10n ) = (A × B) × (10m+n ) .
(7.3 × 105 ) × (6.8 × 104 ) = (7.3 × 6.8) × (105+4 )
(7.3 × 6.8) × (105+4 ) = (49.64) × (109 )
(49.64) × (109 ) = 49.64 × 109
9
1
Write the answer in scientific notation.
9
49.64 × 10 = (4.964 × 10 ) × 10
Apply the law of exponents − add the exponents of the power.
49.64 × 109 = 4.964 × 1010
(7.3 × 105 ) × (6.8 × 104 ) = 4.964 × 1010
iv) (4.8 × 109 ) ÷ (5.79 × 107 )
(4.8 × 109 ) ÷ (5.79 × 107 )
Apply the rule (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n ) .
(4.8 × 109 ) ÷ (5.79 × 107 ) = (4.8 ÷ 5.79) × 109−7 Apply the law of exponents − subtract the exponents of the power.
(4.8 ÷ 5.79) × 109−7 = (0.829) × 102
2
−1
2
(0.829) × 10 = (8.29 × 10 ) × 10
Write the answer in scientific notation.
Apply the law of exponents − add the exponents of the power.
(8.29 × 10−1 ) × 102 = 8.29 × 101
Concept Problem Revisited
The distance between the sun and Neptune would be written as 4.5 × 109 km and the diameter of an electron would
be written as 2.2 × 10−13 in.
Vocabulary
Scientific Notation
Scientific notation is a way of writing numbers in the form of a number between 1 and 10 multiplied by a
power of 10. The number 196.5 written in scientific notation is 1.965 × 102 and the number 0.0760 written in
scientific notation is 7.60 × 10−2 .
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Chapter 4. Exponents
Guided Practice
1. Express the following product in scientific notation: (4 × 1012 )(9.2 × 107 )
2. Express the following quotient in scientific notation:
6,400,000
0.008
3. If a = 0.000415, b = 521, and c = 71, 640, find an approximate value for
notation.
ab
c .
Express the answer in scientific
Answers:
1. Apply the rule
(A × 10m ) × (B × 10n ) = (A × B) × (10m+n )
(4 × 1012 ) × (9.2 × 107 ) = (4 × 9.2) × (1012+7 )
(4 × 9.2) × (1012+7 ) = 36.8 × 1019
Express the answer in scientific notation.
36.8 × 1019 = (3.68 × 101 ) × 1019
(3.68 × 101 ) × 1019 = 3.68 × 1020
(4 × 1012 )(9.2 × 107 ) = 3.68 × 1020
2. Begin by expressing the numerator and the denominator in scientific notation.
6.4×106
8.0×10−3
Apply the rule
(A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m+n )
.
(6.4 × 106 ) ÷ (8.0 × 10−3 ) = (6.4 ÷ 8.0) × (106−−3 ) Apply the law of exponents − subtract the exponents of the powers.
(6.4 ÷ 8.0) × (106−−3 ) = (0.8) × (109 )
(0.8) × (109 ) = 0.8 × 109
Express the answer in scientific notation.
0.8 × 109 = (8.0 × 10−1 ) × 109
0.8 × 109 = 8.0 × 10−1 × 109
Apply the law of exponents − add the exponents of the powers.
8.0 × 10−1 × 109 = 8.0 × 108
6, 400, 000
= 8.0 × 108
0.008
Express the answer in scientific notation.
3. Express all values in scientific notation.
0.000415 = 4.15 × 10−4
521 = 5.21 × 102
71, 640 = 7.1640 × 104
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4.6. Scientific Notation
www.ck12.org
Use the values in scientific notation to determine an approximate value for
ab
c
=
ab
c .
(4.15×10−4 )(5.21×102 )
7.1640×104
In the numerator, apply the rule
(A × 10m ) × (B × 10n ) = (A × B) × (10m+n )
(4.15 × 10−4 )(5.21 × 102 ) (4.15 × 5.21) × (10−4 × 102 )
=
7.1640 × 104
7.1640 × 104
−4
2
(4.15 × 5.21) × (10 × 10 ) 21.6215 × 10−2
=
7.1640 × 104
7.1640 × 104
Apply the rule (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n ) .
21.6215 × 10−2
= (21.6215 ÷ 7.1640) × (10−2−4 )
7.1640 × 104
(21.6215 ÷ 7.1640) × (10−2 × 104 ) = 3.018 × 10−6
Practice
Express each of the following in scientific notation:
1.
2.
3.
4.
5.
42,000
0.00087
150.64
56,789
0.00947
Express each of the following in standard form:
6.
7.
8.
9.
10.
4.26 × 105
8 × 104
5.967 × 1010
1.482 × 10−6
7.64 × 10−3
Perform the indicated operations and express the answer in scientific notation
11.
12.
13.
14.
8.9 × 104 + 4.3 × 105
8.7 × 10−4 − 6.5 × 10−5
(5.3 × 106 ) × (7.9 × 105 )
(3.9 × 108 ) ÷ (2.8 × 106 )
For the given values, perform the indicated operations for
standard form.
15. .
348
ab
c
and express the answer in scientific notation and
www.ck12.org
Chapter 4. Exponents
a = 76.1
b = 818, 000, 000
c = 0.000016
16. .
a = 9.13 × 109
b = 5.45 × 10−23
c = 1.62
Summary
You learned that in an expression like 2x , the "2" is the base and the "x" is the exponent. You learned the following
laws of exponents that helped you to simplify expressions with exponents:
•
•
•
•
•
•
•
•
am × an = am+n
am
m−n (if m > n, a 6= 0)
an = a
m
n
(a ) = amn
n
n n
(ab)
= aan b
a n
= bn (b 6= 0)
b
a0 = 1 (a 6= 0)
a−m = a1m
√
√ m
m
n
a n = am = n a
You learned that scientific notation is a way to express large or small numbers in the form
x = a × 10n
where
1 ≤ a < 10 and n ∈ Z.
You learned that to solve exponential equations with variables in the exponent you should try to rewrite the equations
so the bases are the same. Then, set the exponents equal to each other and solve. If the equation has a variable in the
base you can try to get rid of the exponent or, make the exponents on each side of the equation the same and then set
the bases equal to each other and solve.
Finally, you learned all about exponential functions. You learned that for exponential functions of the form y =
x
ab c + d, if 0 < b < 1 then the function is decreasing and represents exponential decay. If b > 1 then the function
is increasing and represents exponential growth. Exponential functions are used in many real-life situations such as
with the decay of radioactive isotopes and with interest that compounds.
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C HAPTER
5
Polynomials
Chapter Outline
5.1
A DDITION AND S UBTRACTION OF P OLYNOMIALS
5.2
M ULTIPLICATION OF P OLYNOMIALS
5.3
S PECIAL P RODUCTS OF P OLYNOMIALS
Introduction
Here you’ll learn all about polynomials. You’ll start by learning how to add, subtract, and multiply polynomials.
Then you will learn how to factor polynomials, which can be thought of as the opposite of multiplying. Next you’ll
learn how to divide polynomials and how this connects to factoring. Finally, you’ll learn how to use your graphing
calculator to graph polynomials in order to determine the factors and roots of polynomials.
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Chapter 5. Polynomials
5.1 Addition and Subtraction of Polynomials
Here you’ll learn how to add and subtract polynomials.
You are going to build a rectangular garden in your back yard. The garden is 2 m more than 1.5 times as long as it
is wide. Write an expression to show the area of the garden.
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Guidance
The word polynomial comes from the Greek word poly meaning “many”. Polynomials are made up of one or more
terms and each term must have an exponent that is 0 or a whole number. This means that 3x2 +2x+1 is a polynomial,
but 3x0.5 + 2x−2 + 1 is not a polynomial. Some common polynomials have special names based on how many terms
they have:
• A monomial is a polynomial with just one term. Examples of monomials are 3x, 2x2 and 7.
• A binomial is a polynomial with two terms. Examples of binomials are 2x + 1, 3x2 − 5x and x − 5.
• A trinomial is a polynomial with three terms. An example of a trinomial is 2x2 + 3x − 4.
To add and subtract polynomials you will go through two steps.
1. Use the distributive property to remove parentheses. Remember that when there is no number in front of the
parentheses, it is like there is a 1 in front of the parentheses. Pay attention to whether or not the sign in front
of the parentheses is + or −, because this will tell you if the number you need to distribute is +1 or −1.
2. Combine similar terms. This means, combine the x2 terms with the x2 terms, the x terms with the x terms, etc.
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Example A
Find the sum: (3x2 + 2x − 7) + (5x2 − 3x + 3).
Solution: First you want to remove the parentheses. Because this is an addition problem, it is like there is a +1 in
front of each set of parentheses. When you distribute a +1, none of the terms will change.
1(3x2 + 2x − 7) + 1(5x2 − 3x + 3) = 3x2 + 2x − 7 + 5x2 − 3x + 3
Next, combine the similar terms. Sometimes it can help to first reorder the expression to put the similar terms next
to one another. Remember to keep the signs with the correct terms. For example, in this problem the 7 is negative
and the 3x is negative.
3x2 + 2x − 7 + 5x2 − 3x + 3 = 3x2 + 5x2 + 2x − 3x − 7 + 3
= 8x2 − x − 4
This is your final answer.
Example B
Find the difference: (5x2 + 8x + 6) − (4x2 + 5x + 4).
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a −1
in front of the second set of parentheses. When you distribute a −1, each term inside that set of parentheses will
change its sign.
1(5x2 + 8x + 6) − 1(4x2 + 5x + 4) = 5x2 + 8x + 6 − 4x2 − 5x − 4
Next, combine the similar terms. Remember to keep the signs with the correct terms.
5x2 + 8x + 6 − 4x2 − 5x − 4 = 5x2 − 4x2 + 8x − 5x + 6 − 4
= x2 + 3x + 2
This is your final answer.
Example C
Find the difference: (3x3 + 6x2 − 7x + 5) − (4x2 + 3x − 8)
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a −1
in front of the second set of parentheses. When you distribute a −1, each term inside that set of parentheses will
change its sign..
1(3x3 + 6x2 − 7x + 5) − 1(4x2 + 3x − 8) = 3x3 + 6x2 − 7x + 5 − 4x2 − 3x + 8
Next, combine the similar terms. Remember to keep the signs with the correct terms.
3x3 + 6x2 − 7x + 5 − 4x2 − 3x + 8 = 3x3 + 6x2 − 4x2 − 7x − 3x + 5 + 8
= 3x3 + 2x2 − 10x + 13
This is your final answer.
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Chapter 5. Polynomials
Concept Problem Revisited
Remember that the area of a rectangle is length times width.
Area = l × w
Area = (1.5x + 2)x
Area = 1.5x2 + 2x
Vocabulary
Binomial
A binomial has two terms that are added or subtracted from each other. Each of the terms of a binomial is a
variable (x), a product of a number and a variable (4x), or the product of multiple variables with or without a
number (4x2 y + 3). One of the terms in the binomial can be a number.
Monomial
A monomial can be a number or a variable (like x) or can be the product of a number and a variable (like 3x
or 3x2 ). A monomial has only one term.
Polynomial
A polynomial, by definition, is also a monomial or the sum of a number of monomials. So 3x2 can be
considered a polynomial, 2x + 3 can be considered a polynomial, and 2x2 + 3x − 4 can be considered a
polynomial.
Trinomial
A trinomial has three terms (4x2 + 3x − 7). The terms of a trinomial can be a variable (x), a product of a
number and a variable (3x), or the product of multiple variables with or without a number (4x2 ). One of the
terms in the trinomial can be a number (−7).
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often
referred to as the literal coefficient.
Guided Practice
1. Find the sum: (2x2 + 4x + 3) + (x2 − 3x − 2).
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5.1. Addition and Subtraction of Polynomials
2. Find the difference: (5x2 − 9x + 7) − (3x2 − 5x + 6).
3. Find the sum: (8x3 + 5x2 − 4x + 2) + (4x3 + 7x − 5).
Answers:
1. (2x2 + 4x + 3) + (x2 − 3x − 2) = 2x2 + 4x + 3 + x2 − 3x − 2 = 3x2 + x + 1
2. (5x2 − 9x + 7) − (3x2 − 5x + 6) = 5x2 − 9x + 7 − 3x2 + 5x − 6 = 2x2 − 4x + 1
3. (8x3 + 5x2 − 4x + 2) + (4x3 + 7x − 5) = 8x3 + 5x2 − 4x + 2 + 4x3 + 7x − 5 = 12x3 + 5x2 + 3x − 3
Practice
For each problem, find the sum or difference.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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(x2 + 4x + 5) + (2x2 + 3x + 7)
(2r2 + 6r + 7) − (3r2 + 5r + 8)
(3t 2 − 2t + 4) + (2t 2 + 5t − 3)
(4s2 − 2s − 3) − (5s2 + 7s − 6)
(5y2 + 7y − 3) + (−2y2 − 5y + 6)
(6x2 + 36x + 13) − (4x2 + 13x + 33)
(12a2 + 13a + 7) + (9a2 + 15a + 8)
(9y2 − 17y − 12) + (5y2 + 12y + 4)
(11b2 + 7b − 12) − (15b2 − 19b − 21)
(25x2 + 17x − 23) − (−14x3 − 14x − 11)
(−3y2 + 10y − 5) − (5y2 + 5y + 8)
(−7x2 − 5x + 11) + (5x2 + 4x − 9)
(9a3 − 2a2 + 7) + (3a2 + 8a − 4)
(3x2 − 2x + 4) − (x2 + x − 6)
(4s3 + 4s2 − 5s − 2) − (−2s2 − 5s + 6)
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Chapter 5. Polynomials
5.2 Multiplication of Polynomials
Here you will learn how to multiply polynomials using the distributive property.
Jack was asked to frame a picture. He was told that the width of the frame was to be 5 inches longer than the glass
width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass and finds
the height to width ratio is 4:3. Write the expression to determine the area of the picture frame.
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Guidance
To multiply polynomials you will need to use the distributive property. Recall that the distributive property says that
if you start with an expression like 3(5x + 2), you can simplify it by multiplying both terms inside the parentheses
by 3 to get a final answer of 15x + 6.
When multiplying polynomials, you will need to use the distributive property more than once for each problem.
Example A
Find the product: (x + 6)(x + 5)
Solution: To answer this question you will use the distributive property. The distributive property would tell you to
multiply x in the first set of parentheses by everything inside the second set of parentheses , then multiply 6 in the
first set of parentheses by everything in the second set of parentheses . Here is what that looks like:
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5.2. Multiplication of Polynomials
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Example B
Find the product: (2x + 5)(x − 3)
Solution: Again, use the distributive property. The distributive property tells you to multiply 2x in the first set of
parentheses by everything inside the second set of parentheses , then multiply 5 in the first set of parentheses by
everything in the second set of parentheses . Here is what that looks like:
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Chapter 5. Polynomials
Example C
Find the product: (4x + 3)(2x2 + 3x − 5)
Solution: Even though at first this question may seem different, you can still use the distributive property to find
the product. The distributive property tells you to multiply 4x in the first set of parentheses by everything inside
the second set of parentheses, then multiply 3 in the first set of parentheses by everything in the second set of
parentheses. Here is what that looks like:
Concept Problem Revisited
Jack was asked to frame a picture. He was told that the width of the frame was to be 5 inches longer than the glass
width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass and finds
the height to width ratio is 4:3. Write the expression to determine the area of the picture frame.
What is known?
• The width is 5 inches longer than the glass
• The height is 7 inches longer than the glass
• The glass has a height to width ratio of 4:3
The equations:
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5.2. Multiplication of Polynomials
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• The height of the picture frame is 4x + 7
• The width of the picture frame is 3x + 5
The formula:
Area = w × h
Area = (3x + 5)(4x + 7)
Area = 12x2 + 21x + 20x + 35
Area = 12x2 + 41x + 35
Vocabulary
Distributive Property
The distributive property states that the product of a number and a sum is equal to the sum of the individual
2
products of the number and the addends.
For example, in the expression: 3 (x + 5), the distributive property
2
states that the
product of a number 3 and a sum (x + 5) is equal to the sum of the individual products of the
2
number 3 and the addends (x and 5).
Like Terms
Like terms refers to terms in which the degrees match and the variables match. For example 3x and 4x are like
terms. Like terms are also known as similar terms.
Guided Practice
1. Find the product: (x + 3)(x − 6)
2. Find the product: (2x + 5)(3x2 − 2x − 7)
3. An average football field has the dimensions of 160 ft by 360 ft. If the expressions to find these dimensions were
(3x + 7) and (7x + 3), what value of x would give the dimensions of the football field?
Answers:
1. (x + 3)(x − 6)
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Chapter 5. Polynomials
2. (2x + 5)(3x2 − 2x − 7)
3. Area = l × w
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5.2. Multiplication of Polynomials
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Area = 360 × 160
(7x + 3) = 360
7x = 360 − 3
7x = 357
x = 51
(3x + 7) = 160
3x = 160 − 7
3x = 153
x = 51
The value of x that satisfies these expressions is 51.
Practice
Use the distributive property to find the product of each of the following polynomials:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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(x + 4)(x + 6)
(x + 3)(x + 5)
(x + 7)(x − 8)
(x − 9)(x − 5)
(x − 4)(x − 7)
(x + 3)(x2 + x + 5)
(x + 7)(x2 − 3x + 6)
(2x + 5)(x2 − 8x + 3)
(2x − 3)(3x2 + 7x + 6)
(5x − 4)(4x2 − 8x + 5)
9a2 (6a3 + 3a + 7)
−4s2 (3s3 + 7s2 + 11)
(x + 5)(5x3 + 2x2 + 3x + 9)
(t − 3)(6t 3 + 11t 2 + 22)
(2g − 5)(3g3 + 9g2 + 7g + 12)
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Chapter 5. Polynomials
5.3 Special Products of Polynomials
Here you will learn about special cases of binomial multiplication.
A flower is homozygous blue (RR) and another flower is homozygous white (rr). Use a Punnett square to show that
a mixture of the two can produce a white flower.
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Guidance
There are two special cases of multiplying binomials. If you can learn to recognize them, you can multiply these
binomials more quickly.
Here are the two special products that you should learn to recognize:
Special Case 1 (Binomial Squared): (x ± y)2 = x2 ± 2xy + y2
• Example: (x + 5)2 = x2 + 10x + 25
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5.3. Special Products of Polynomials
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• Example: (2x − 8)2 = 4x2 − 32x + 64
Special Case 2 (Difference of Perfect Squares): (x + y)(x − y) = x2 − y2
• Example: (5x + 10)(5x − 10) = 25x2 − 100
• Example: (2x − 4)(2x + 4) = 4x2 − 16
Keep in mind that you can always use the distributive property to do the multiplications if you don’t notice that the
problem is a special case.
Example A
Find the product: (x + 11)2
Solution: This is an example of Special Case 1. You can use that pattern to quickly multiply.
(x + 11)2 = x2 + 2 · x · 11 + 112
= x2 + 22x + 121
You can verify that this is the correct answer by using the distributive property:
Example B
Find the product: (x − 7)2
Solution: This is another example of Special Case 1. You can use that pattern to quickly multiply.
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Chapter 5. Polynomials
(x − 7)2 = x2 − 2 · x · 7 + 72
= x2 − 14x + 49
You can verify that this is the correct answer by using the distributive property:
Example C
Find the product: (x + 9)(x − 9)
Solution: This is an example of Special Case 2. You can use that pattern to quickly multiply.
(x + 9)(x − 9) = x2 − 92
= x2 − 81
You can verify that this is the correct answer by using the distributive property:
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5.3. Special Products of Polynomials
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Concept Problem Revisited
Each flower will have one-half of the blue genes and one-half of the white genes. Therefore the equation formed
will be:
0.5B + 0.5W
The offspring will have the genetic makeup (the mixture produced) using the equation:
(0.5B + 0.5W )2
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Chapter 5. Polynomials
Notice that this is an example of Special Case 1. You can expand the offspring genetic makeup equation to find out
the percentage of offspring (or flowers) that will be blue, white, or light blue.
Therefore 25% of the offspring flowers will be blue, 50% will be light blue, and 25% will be white.
Vocabulary
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number
and a sum is equal to the sum of the individual products of the number and the addends. For example, in
the expression: 3(x + 5), the distributive property states that the product of a number (3) and a sum (x + 5) is
equal to the sum of the individual products of the number (3) and the addends (x and 5).
Guided Practice
1. Expand the following binomial: (x + 4)2 .
2. Expand the following binomial: (5x − 3)2 .
3. Determine whether or not each of the following is a difference of two perfect squares:
a) a2 − 16
b) 9b2 − 49
c) c2 − 60
Answers:
1. (x + 4)2 = x2 + 4x + 4x + 16 = x2 + 8x + 16.
2. (5x − 3)2 = 25x2 − 15x − 15x + 9 = 25x2 − 30x + 9
3. a)Yes, a2 − 16 = (a + 4)(a − 4)
b) Yes, 9b2 − 49 = (3b + 7)(3b − 7)
c) No, 60 is not a perfect square.
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5.3. Special Products of Polynomials
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Practice
Expand the following binomials:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
(t + 12)2
(w + 15)2
(2e + 7)2
(3z + 2)2
(7m + 6)2
(g − 6)2
(d − 15)2
(4x − 3)2
(2p − 5)2
(6t − 7)2
Find the product of the following binomials:
11.
12.
13.
14.
15.
(x + 13)(x − 13)
(x + 6)(x − 6)
(2x + 5)(2x − 5)
(3x + 4)(3x − 4)
(6x + 7)(6x − 7)
Summary
You learned that adding, subtracting, and multiplying polynomials all rely on the distributive property. You also
learned that factoring is the reverse of multiplying because when you factor a polynomial you are trying to rewrite
the polynomial as a product of other polynomials. You learned how to factor completely by first looking for common
factors and then using other methods to factor the remaining expression. You learned special cases of factoring to
watch out for including the difference of perfect squares, perfect square trinomials, and the sum and difference of
cubes. You learned how factoring can allow you to solve a quadratic equation with the help of the zero product
property.
You learned how to divide polynomials and how dividing polynomials can connect to factoring. You learned about
the factor theorem and that polynomial long division can actually help you to factor higher degree polynomials.
Finally, you learned that factors of polynomials connect to their x-intercepts, and that you can use a graphing
calculator to graph polynomial functions in order to find out more information about the factors of the polynomial.
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